and we want to find the shortest distance from a point P with position vector (3i + j + 16k) to this plane.

First, find the perpendicular distance of the plane from the origin. To do this, consider the position vector r of an arbitrary point on the plane. Now imagine a vector that is perpendicular to the plane and passes through the origin. (I would recommend drawing a diagram.) It is now evident that the vector r, that is the vector from the origin to the arbitrary point on the plane, is the hypotenuse of a triangle with its base on the plane, and the remaining sideformed by the perpendicular vector. This perpendicular vector is called n.

Looking at the triangle, we can see that the perpendicular distance of the plane from the origin, call it OA, is |r|cos θ.
Now, using the dot product:

From the equation of the plane given at the beginning, we know that ρ = 13. We also know that n = 2i - j - 3k. Therefore |n| = sqrt(4 + 1 + 9) = sqrt(14).

Hence, OA = 13/sqrt(14).

Now we need to form a plane containing the point P that is parallel to our original plane. As the planes are parallel, the vector n must be the same for both of them. If we let a equal the position vector of P, we have the equation of the plane:

r.(2i - j - 3k) = (3i + j + 16k).(2i - j - 3k) = 6-1-48 = -43.

What we need to do now is the same that we did for our original plane, but for our new plane this time. We can use the result from before, so if we say that OB is the perpendicular distance of this plane to the origin:

OB = ρ/|n| = -43/sqrt(14)

The fact that this value is negative indicates that P is on the other side of the origin to the original plane. Keeping this in mind, we can see that to get the distance between the original plane and P, we need to add the magnitudes of the two numbers together.
Hence, if the perpendicular distance of P from the original plane is d: